Find the length of the common chord of the el | vedclass

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(A) The equation of the ellipse is $\frac{(x - 2)^2}{3^2} + \frac{(y + 2)^2}{2^2} = 1$.The equation of the circle is $x^2 - 4x + 4 + y^2 + 2y + 1 = -4

Vedclass · Accepted Answer

$0$

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(A) The equation of the ellipse is $\frac{(x - 2)^2}{3^2} + \frac{(y + 2)^2}{2^2} = 1$.The equation of the circle is $x^2 - 4x + 4 + y^2 + 2y + 1 = -4 + 4 + 1$,which simplifies to $(x - 2)^2 + (y + 1)^2 = 1$.Let $X = x - 2$ and $Y = y + 2$. Then the ellipse is $\frac{X^2}{9} + \frac{Y^2}{4} = 1$.The circle is $(X)^2 + (Y - 1)^2 = 1$,or $X^2 + Y^2 - 2Y = 0$.Substituting $X^2 = 9(1 - \frac{Y^2}{4})$ into the circle equation: $9 - \frac{9Y^2}{4} + Y^2 - 2Y = 0$.$9 - \frac{5Y^2}{4} - 2Y = 0 \implies 5Y^2 + 8Y - 36 = 0$.Solving for $Y$: $Y = \frac{-8 \pm \sqrt{64 - 4(5)(-36)}}{10} = \frac{-8 \pm \sqrt{64 + 720}}{10} = \frac{-8 \pm \sqrt{784}}{10} = \frac{-8 \pm 28}{10}$.$Y = 2$ or $Y = -3.6$. Since the ellipse is $\frac{Y^2}{4} \le 1$,$Y$ must be in $[-2, 2]$.Thus $Y = 2$ is the only solution,which gives $X = 0$.At $X = 0, Y = 2$,we have $x = 2$ and $y = 0$. The curves touch at $(2, 0)$.Since they touch at a single point,the length of the common chord is $0$.

Find the length of the common chord of the ellipse $\frac{(x - 2)^2}{9} + \frac{(y + 2)^2}{4} = 1$ and the circle $x^2 + y^2 - 4x + 2y + 4 = 0$.

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